Stability of the SIR epidemic model — Jacobian is singular

Question

The SIR epidemic model presents three differential equations for three time-dependent variables — $S(t)$, $I(t)$, $R(t)$.

\begin{aligned} \frac{dS}{dt} &= -\beta SI\\ \frac{dI}{dt} &= \beta SI - \gamma I\\ \frac{dR}{dt} &= \gamma I\\ \end{aligned}

For the SIR model, I calculate the Jacobian matrix $J$, as follows

\begin{equation*} J = \begin{pmatrix} \frac{\partial S}{\partial S} & \frac{\partial S}{\partial I} & \frac{\partial S}{\partial R} \\ \frac{\partial I}{\partial S} & \frac{\partial I}{\partial I} & \frac{\partial I}{\partial R} \\ \frac{\partial R}{\partial S} & \frac{\partial R}{\partial I} & \frac{\partial R}{\partial R} \end{pmatrix} = \begin{pmatrix} - \beta I & - \beta S & 0 \\ \beta I & \beta S -\gamma & 0 \\ 0 & \gamma & 0 \end{pmatrix} \end{equation*}

In this case, the characteristic polynomial is given by

$$(-\lambda) \left( \lambda^2+(\beta I-\beta S+\gamma)\lambda+\beta I\gamma \right)$$

To determine the stability of the disease-free equilibrium we substitute $S = 1$ and $I = 0$ for $S$ and $I$. Hence, the characteristic polynomial becomes

$$-\lambda^2(\lambda-\beta+\gamma)$$

This has three roots: $\lambda_1$, $\lambda_2 =0$ and $\lambda_3=\beta-\gamma$. The last eigenvalue to be negative is required that $\beta<\gamma$. How can I determine the stability if two of the three eigenvalues are $0$? I have seen some solutions for $2 \times 2$ matrix, but I don't understand them. Is there any easy reason for being stable or unstable?

Answer 1

You do not need any Jacobians here. First note that the third equation is separate from the other two, hence it can be ignored (mathematically, you have $S+R+I=1$, so the variables are not independent). Second, look at the second equation: $$ \dot I=\beta SI-\gamma I=I(\beta S-\gamma). $$ We clearly have the growth for $I$ if $\beta S>\gamma$, hence at the initial time moment, when $S=1$, we get the condition that we have epidemics (the disease free equilibrium is unstable) if $\beta>\gamma$, which is equivalent to what you have found.

Answer 2

But in other cases, what to do if one or more eigenvalues be zero? Like the SIS model case with constant population.